Let $f \in \mathbb{Z}[x]$ be a polynomial of degree $k \geq 1$ with first coefficient $a_k > 0$, $\mathcal{A} = f(\mathbb{N}) \cap \mathbb{N}^*$, $m \in \mathbb{N}^*$ odd and $\mathcal{P}$ a set of prime numbers such that there exists $\lambda \in \mathbb{R}^+$ with $$ \sum_{p \in \mathcal{P}(z)} \frac{1}{p} \sim \lambda \ln(\ln z), $$ where $\mathcal{P}(z)=\mathcal{P}\cap[1,z]$. Fixing $\varepsilon > 0$ and $z = e^{(\ln x(2(1+\varepsilon)\lambda e k\ln (\ln x)^{-1}))}$, for $x\rightarrow + \infty$ it holds $$ \vert \{ n \leq x : p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z) \} \vert \leq x \prod_{p \in \mathcal{P}(z)} \Big( 1 - \frac{\rho(p)}{p} \Big) + O \Big( \frac{x}{(\ln z)^{2k\lambda}} \Big). $$
I understood the proof except for the following inequality:
$$\vert \{ n \leq x : p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z) \} \vert < z + \mathcal{S}(\mathcal{A},f(x),M(z)),$$
with $\mathcal{S}(\mathcal{A},f(x),M(z)) = \vert \{ f(n) \in \mathcal{A} \cap [1,f(x)] : (f(n),M(z))=1 \} \vert$ and $M(z)=\prod_{p \in \mathcal{P}(z)} p$
I know that the condition $(f(n),M(z))=1$ is equal to $(p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z))$ and that, for $x$ large enough $n \leq x$ is equal to $f(n) \leq f(x)$.
The fact is that I have no hypothesis on the function $f$, so I don't know how to proceed. I think it could assume negative values or that it could be non injective. So it holds, of course, $\vert \{ n \leq x : p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z) \} \vert \geq \mathcal{S}(\mathcal{A},f(x),M(z))$.
I thought about counting the elements that are in $\vert \{ n \leq x : p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z) \} \vert$ but not in $\mathcal{S}(\mathcal{A},f(x),M(z))$, but again, with no hypothesis on the $f$, i have no idea how to do it.
Edit: I think that, since for $x\rightarrow + \infty$ also $z \rightarrow + \infty$, as I already said, $n \leq x$ is equivalent to $f(n) \leq f(x)$. This means that $$\vert \{ z < n \leq x : p \vert f(n) \Rightarrow p \not\in \mathcal{P}(z) \} \vert \leq \mathcal{S}(\mathcal{A},f(x),M(z)),$$ whilst for the $n \leq z$, I simply increase the set with $z$. It seems that it works, but I'm not totally sure about it.